How to get filename of the __main__ module in Python?

Question

Suppose I have two modules:

a.py:

import b print __name__, __file__ 

b.py:

print __name__, __file__ 

I run the "a.py" file. This prints:

b        C:\path\to\code\b.py __main__ C:\path\to\code\a.py 

Question: how do I obtain the path to the __main__ module ("a.py" in this case) from within the "b.py" library?

Answer 1

import __main__ print(__main__.__file__) 

Answer 2

Perhaps this will do the trick:

import sys from os import path print(path.abspath(str(sys.modules['__main__'].__file__))) 

Note that, for safety, you should check whether the __main__ module has a __file__ attribute. If it's dynamically created, or is just being run in the interactive python console, it won't have a __file__:

python >>> import sys >>> print(str(sys.modules['__main__'])) <module '__main__' (built-in)> >>> print(str(sys.modules['__main__'].__file__)) AttributeError: 'module' object has no attribute '__file__' 

A simple hasattr() check will do the trick to guard against scenario 2 if that's a possibility in your app.